# Volume of Sphere/Proof by Cavalieri

## Theorem

The volume $V$ of a sphere of radius $r$ is given by:

- $V = \dfrac {4 \pi r^3} 3$

## Proof

Consider a sphere $S$ of radius $r$.

Consider a right circular cylinder $C$ whose bases are circles of radius $r$ and whose height is $2 r$.

Let a double napped cone $K$ each of whose nappes has bases which coincide with the bases of $C$.

Let $K$ be removed from $C$ to leave a solid figure $C'$ described as a right circular cylinder with a conical hollow at either end.

The bases of $C$ are circles of radius $r$.

From Area of Circle, the area of each base is therefore $\pi r^2$.

From Volume of Right Circular Cylinder, $C$ has volume given by:

\(\ds V_C\) | \(=\) | \(\ds \pi r^2 \times 2 r\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \pi r^3\) |

$K$ has twice the volume of each of its nappes.

Each of the nappes of $K$ is of height $r$ and has a base of area $\pi r^2$.

Hence from Volume of Cone, $K$ has volume given by:

\(\ds V_K\) | \(=\) | \(\ds 2 \times \dfrac 1 3 \times \pi r^2 \times r\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac 2 3 \pi r^3\) |

Thus the volume $V_{C'}$ of $C'$ is given by:

\(\ds V_{C'}\) | \(=\) | \(\ds 2 \pi r^3 - \dfrac 2 3 \pi r^3\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {\left({6 - 2}\right) \pi r^3} 3\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {4 \pi r^3} 3\) |

Let the sphere $S$ be positioned so as to be tangent to the plane containing the base of $C$

Consider a plane parallel to the base of $C$ intersecting both $S$ and $C$ at a distance $x$ from the center of $S$.

Let $y$ be the radius of the circle which is the intersection of $S$ with that plane.

From Pythagoras's Theorem, $y$ is given by:

- $x^2 + y^2 = r^2$

and so:

- $y^2 = r^2 - x^2$

Thus the area of this circle is given by:

- $A = \pi \left({r^2 - x^2}\right)$

Similarly, consider the intersection of $C'$ with the same plane.

This consists of:

with:

- the circle of radius $x$ which is the intersection of the plane with the cone

removed from it.

So the area of the remaining circle with the inner circle removed is:

- $A' = \pi r^2 - \pi x^2$

from Area of Circle.

Thus the intersection of the plane with both $S$ and $C'$ are the same.

Thus by Cavalieri's Principle $S$ and $C'$ have the same volume.

Hence the result.

$\blacksquare$

## Historical Note

This proof was given by Bonaventura Francesco Cavalieri.

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.14$: Cavalieri ($\text {1598}$ – $\text {1647}$)